# physic

A ball is thrown straight upward and returns to the thrower's hand after 2.00 s in the air. A second ball thrown at an angle of 40.0° with the horizontal reaches the same maximum height as the first ball.
(a) At what speed was the first ball thrown?
m/s

(b) At what speed was the second ball thrown?
m/s

asked by mark
1. Tr = Rise time, Tf = Fall time.

a. Tr + Tf = 2 s, Tf = Tr, Tr+Tr = 2, 2Tr = 2, Tr = 1 s.

V = Vo + g*Tr = 0, Vo = -g*Tr =
9.8*1 = 9.8 m/s.

b. Yo = 9.8 m/s(Part a), Vo*sin40
= 9.8, Vo = 9.8/sin40 = 15.25 m/s.

posted by Henry

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